If $x \star y = 2x^{2}-y^{2}$ and $x \veebar y = 3x-3$, find $-4 \star (2 \veebar 0)$.
Solution: First, find $2 \veebar 0$ $ 2 \veebar 0 = (3)(2)-3$ $ \hphantom{2 \veebar 0} = 3$ Now, find $-4 \star 3$ $ -4 \star 3 = 2(-4)^{2}-3^{2}$ $ \hphantom{-4 \star 3} = 23$.